Consider a $n$ by $n$ by $n$ grid represented by the set of $3$-uples $S=\{1,2,\dots, n\}^3$.  
A line (resp. slice) of $S$ is a subset of cardinal $n$ (resp. $n^2$) where two components (resp. one component) of the $3$-uples are fixed.  

Some boxes of the grid are black (and the remaining ones are white), they are represented by a subset $B \subseteq S$. For a fixed $r \le n$, consider the following assumptions on $B$:  

- every line contains exactly $r$ black boxes.
- in every slice, the black boxes can be filled collinearly, i.e. there is an ordering $b_1, b_2, \dots, b_{rn}$ such that for all $i>1$ there is $j<i$ such that $b_i$ and $b_j$ are in a same line.      

*Example*: For $r=2$ and $n = 4$, the following combination (given by the pictures of $4$ parallel slices) satisfies the two above assumptions.


$\substack{
\displaystyle{◻◻◼◼} \cr
\displaystyle{◼◼◻◻} \cr
\displaystyle{◼◻◼◻} \cr
\displaystyle{◻◼◻◼} 
} $
$\substack{
\displaystyle{◻  ◼  ◼  ◻} \cr
\displaystyle{◼  ◻  ◼  ◻} \cr
\displaystyle{◻  ◼  ◻  ◼} \cr
\displaystyle{◼  ◻  ◻  ◼} 
} $
$\substack{
\displaystyle{◼  ◻  ◻  ◼} \cr
\displaystyle{◻  ◻  ◼  ◼} \cr
\displaystyle{◼  ◼  ◻  ◻} \cr
\displaystyle{◻  ◼  ◼  ◻} 
} $
$\substack{
\displaystyle{◼  ◼  ◻  ◻} \cr
\displaystyle{◻  ◼  ◻  ◼} \cr
\displaystyle{◻  ◻  ◼  ◼} \cr
\displaystyle{◼  ◻  ◼  ◻} 
} $ 

An ordering $b_1, b_2, \dots, b_{rn^2}$ of $B$ is Eulerian if for all $i>1$ and for all $j<i$, there is $k<i$ such that $b_i$ and $b_k$ are in a same line $l$, and if $b_i$ and $b_j$ are in a same slice $s$, then $l \subset s$.  
Note that the notion of Eulerian ordering is related to the notion of [shelling][1], as explained in [this post][2].  

*Remark*: If $r=n$ then the grid contains black boxes only and the lexicographic ordering is Eulerian.   

**Question**: Has $B$ (satisfying the two above assumptions) an Eulerian ordering if $r \ge 3$?  
 
*Remark*: The above combination (with $r=2$ and $n = 4$) admits no Eulerian ordering (proved below by brute-force search; a conceptual proof would be useful). Note that any *partial* Eulerian ordering of it has length at most $11$.

____
Brute-force search with SAGE  

Computation:
 

    sage: %attach SAGE/EulerianGrid.spyx
    Compiling ./SAGE/EulerianGrid.spyx... 
    sage: B=[[1,1,2],[1,1,4],[1,2,1],[1,2,3],[1,3,1],[1,3,2],[1,4,3],[1,4,4],[2,1,1],[2,1,4],[2,2,2],[2,2,4],[2,3,1],[2,3,3],[2,4,2],[2,4,3],[3,1,2],[3,1,3],[3,2,1],[3,2,2],[3,3,3],[3,3,4],[3,4,1],[3,4,4],[4,1,1],[4,1,3],[4,2,3],[4,2,4],[4,3,2],[4,3,4],[4,4,1],[4,4,2]]
    sage: %time PartialOrdering(B,[],11)
    CPU times: user 29min 4s, sys: 796 ms, total: 29min 5s
    Wall time: 29min 27s


Code:

    # %attach SAGE/EulerianGrid.spyx
    
    from sage.all import *

    cpdef JoinDegree(list L1, list L2):
    	cdef int i,c
    	c=0
    	for i in range(3):
    		if L1[i]==L2[i]:
    			c+=1
    	return c
    
    cpdef IsCollinearList(list l, list L):
    	cdef list i
    	if L==[]:
    		return True
    	for i in L:
    		if JoinDegree(l,i)==2:
    			return True
    	return False
    
    cpdef PartialOrdering(list L, list P, int A):
    	cdef int c,cc
    	cdef list i,j,k,t,LL,PP
    	if len(P)>A:
    		print(P)
    	if L<>[]:
    		for i in L:
    			if IsCollinearList(i,P):
    				cc=0
    				for j in P:
    					if JoinDegree(i,j)==1:
    						c=0
    						for k in P:
    							if JoinDegree(i,k)==2 and JoinDegree(j,k)>=1:
    								c=1
    								break 
    					if c==0:
    						cc=1
    				if cc==0:
    					LL=[t for t in L]
    					PP=[t for t in P]
    					LL.remove(i)
    					PP.append(i)
    					if LL<>[]:
    						PartialOrdering(LL,PP,A)
    					else:
    						return PP


  [1]: https://en.wikipedia.org/wiki/Shelling_(topology)
  [2]: https://mathoverflow.net/q/296273/34538